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11Alexandr11 [23.1K]
3 years ago
15

Calculate the number of moles in 5.96 g KOH

Chemistry
2 answers:
Darina [25.2K]3 years ago
8 0
Collection of data mass(m) =5,96 (given) Molarmass(M) =1(39)+1(16)+1(1)=56 (we use the periodic table to calculate the molar mass of KOH) Moles(n) =?? (It is wat we have to calculate) We can use the formula n=m/M Therefore n=5,96/56=0,10643moles
Bond [772]3 years ago
4 0
? moles KOH = 5.96 x (molar mass of KOH/1 mol)
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What can you tell by tracking oxidation numbers?
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<u>Answer</u>:

By tracking oxidation numbers we can identify the number electron in the atom

<u>Explanation</u>:

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Which type of river is similar to a mature river, but flows more slowly so has less power to change the landscape? Select one: a
Ede4ka [16]

Answer:

The correct option is;

C. Old Age River

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3 years ago
What is typically displayed on the y-axis of a solubility curve
tekilochka [14]
<span>Grams of solute per 100 grams of water</span>
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Read 2 more answers
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baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

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Answer:

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So, ClO₄⁻ will be more strongly hydrated.

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